想請問紅色的地方要怎麼變藍色的

EXAMPLE 3 Solution We differentiate the equation implicitly. (AS) ² = x² + sin.xy d VIR SY d (²) - (x²)+(sin xy) = dx dx Find dy/dx if y² = x² + sinxy (Figure 3.31). d dx dy dx 2y = d 2x + (cos xy) (xy) dx ansa to along or see anuar Differentiate both sides with respect to x... ... treating y as a function of x and using the Chain Rule.

請問有人知道怎麼算嗎？拜託了🙏

11. [-/6 Points] DETAILS Find the constants a and b such that the function is continuous on the entire real line. a = b = LARCALC12M 1.4.073. 5, f(x) = ax + b, -5, 13 X ≤-2 -2<x<3 X23

如圖，為什麼將e^(-2y)積分後會變成-1/2（e^-2y)，求過程

dv=e²² dy v=-10²2²2y

Plzzz🥹 有沒有機率小天才

5.9 若已知臺灣地區的勞工中,男性、女性各佔60%及40%。 ①若已知男性勞工中,外籍勞工佔5%;女性勞工中,外籍勞工佔2%, 那麼從勞工中隨機抽取一人,其爲外籍勞工的機率為何? Nee 2 ②若知該被抽取中者爲外籍勞工,則爲男性的機率為何? 台突围(waivbA

拜託ㄌ🥺

COST: 5.8 設已婚男性看「正晶限時批」的機率是 0.6,而已婚女性看此節目的機率 是 0.5。若已知太太看該節目,而丈夫也看的機率爲 0.8,試求夫婦二人 Pa A 中,至少有一人看該節目的機率。 "Cichert

我想問問D為什麼不能成為樣本資料 謝謝🥹

-1 選擇題 (每小題2 分,共 10 分) (B)1.下列何者是樣本資料? (A) 全國牙醫師的收入分布。 (C) 全國大學生的平均身高。 (B) 選取 200 個高科大學生,調查其每天滑手機時間。 (D) 民國40年到109 年高雄市各區的人口數。 (C)2. 請問下列對於「母體」與「樣本」的定義何者錯誤? (A) 樣本其實就是母體的一部份 (B) 母體就是具有某些相同特質的個體所組成的群體

微積分求解

8. [0/8 Points] Consider the following limit. DETAILS lim Ax-0+ lim Ax-0+ Submit Answer 8 X + Ax Ax Simplify the rational expression as much as possible. 8 x(x + Ax) X 8 X PREVIOUS ANSWERS X Evaluate the one-sided limit. (If an answer does not exist, enter DNE.) DNE LARCALC12M 1.4.029.EP.

請問這題要怎麼算～

1 89. Let f be a function such that f'(x): If 1 + x² 0₁5 2 LIFE AND Qurbliet 1 + x² dig(x) = f(2x + 1), what is g'(x)?sup inslissxs) =

不知道怎麼算(b) 謝謝！

4. 試決定k的值,使f(x)為連續函數: k(x+1) 當æ<3 (a) f(x) = ((b) g(x) = = x² 2 當>3 √2x +5 − √√√x + 6 æ-1 k 呆 k 當T>1 當 æ < 1

想問第4題畫線處 分母(tan²x)-1 =(sin²x/cos²x)-1 =(sin²x-cos²x)/cos²x 為什麼把1/cos²x移到分子以後 分母不是sin²x-cos²x ? 感謝！

x² Example 4 Find lim x 0 sec X 1 SOLUTION The evaluation of this limit requires a little imagination. Since both the numerator and denominator tend to zero as x tends to zero, it is not clear what happens to the fraction. However, we can rewrite the fraction in a more amenable form by multiplying both numerator and denominator by sec x + 1. x2 x² sec x 1 lim x-0 secx 1 - = = x2 sec x 1 - x² (secx + 1) sec² x 1 lim (sin x-0 sin x sec x + 1 sec x + 1) - = Since each of these factors has a limit as x tends to 0, the fraction we began with has a limit: = x² (secx + 1) tan² x - 1 x² cos²x(secx + 1) sin² X 2 (*) ² (cos²x)(secx + 1). 2 ·lim cos²x lim (secx + 1) = (1)(1)(2) = 2. x→0 x ●